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Quiz Number 2 Group 1 – North of Newark Thamer AbuDiak Reynald Benoit Jose Lopez Rosele Lynn Dave Neal Deyanira Pena Professor Kenneth D. Lawerence New Jersey Inst. Of Tech MIS 680

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Problems Assigned Ragsdale/ Dielman Problems Done By: Thamer AbuDiak 5-7/4-11 Reynald Benoit 5-13, 6-18 / 4-1 Jose Lopez 6-6/4-5,5-1 Rosele Lynn 5-10,6-15 Dave Neal 5-16, 6-9 Deyanira Pena 5-19 and 6-12

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Ragsdale 5-7 by Thamer First contract cost Two computer leases are available for Comp-Trail: Company 1: $62,000 initial cost Price of equipment increases by 6% every year Trade in credit: 60% after 1 year. 15% after 2 years. $2,000 Labor cost whenever equipment needs to be replaced. Company 2: $62,000 initial cost Price of equipment increases by 2% every year Trade in credit: 10% after 2 years. 30% after 1 year. $2,000 Labor cost whenever equipment needs to be replaced

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Ragsdale 5-7 by Thamer Set up Decision Variable X 12 : cost of equipment if purchased the 1 st year and replaced the 2 nd year. X 13 : cost of equipment if purchased the 1 st year and replaced the 3 rd year. X 23 : cost of equipment if purchased the 2 nd year and replaced the 3 rd year. X 24 : cost of equipment if purchased the 2 nd year and replaced the 4 th year. X 34 : cost of equipment if purchased the 3 rd year and replaced the 4 th year. X 35 : cost of equipment if purchased the 3 rd year and replaced the 5 th year. X 45 : cost of equipment if purchased the 4 th year and replaced the 5 th year. C 1-7 : Constants. Objective Function Min: C 1 *(X 12 +2000)+ C 2 *(X 13 +2000) + C 3 *(X 23 +2000) + C 4 *(X 24 +2000) +C 5 *(X 34 +2000) + C 6 *(X 35 +2000) + C 7 *(X 45 +2000) Constraints C 1 -C 7 = 0,1. - X 12 – X 13 = -1 } Flow constraint for node 1. X 12 – X 23 – X 24 = 0 } Flow constraint for node 2. X 12 + X 13 – X 34 – X 35 = 0 } Flow constraint for node 3. X 24 + X 34 + X 45 = 0} Flow constraint for node 4. X 35 + X 45 = +1 } Flow constraint for node 5.

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Cost of the first contract Cost of the second contract Ragsdale 5-7 by Thamer Initial Solution Second contract is $5,799 cheaper than the first one

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Cost of the first contract after revision Cost of the second contract after revision Ragsdale 5-7 by Thamer Revised Solution Second contract is $9,799 cheaper than the first one Solver

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Ragsdale 5-7 by Thamer Conclusion The additional $2,000 in labor cost increases the price of the first lease by $8,000 and the second lease by $4,000. The decision of what year to change computers remains the same. Lease 2 remains cheaper than Lease 1. The second contract remains more optimal financially.

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Ragsdale 5-13 by Reynald

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Ragsdale 5-13 by Reynald Set up Decision Variable Xij: tons of products flowing from node i to j Objective Function Min: -24X 13 - 23X 14 -.5X 15 - 25.5X 23 - 25X 24 +.5X 25 - 10X 36 - 11X 47 + 50X 68 + 44X 69 + 45X 610 + 48X 78 + 42X 79 + 43X 710 Constraints X 13 + X 14 + X 15 <= 700 X 23 - X 24 + X 25 <= 500 X 13 + X 23 - X 68 - X 69 - X 610 >= 0 X 14 + X 24 - X 78 - X 79 - X 710 >= 0 350 <= X 13 + X 23 <= 700 300 <= X 14 + X 24 <= 600 X 68 + X 78 = 400 X 69 + X 79 = 300 X 610 + X 710 = 450

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Ragsdale 5-13 by Reynald Excel

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Ragsdale 5-13 by Reynald Conclusion The cotton grower should ship 250 from Statesboro to Claxton 450 from Statesboro to Millen 450 from Brooklet to Claxton 250 from Claxton to Savannah 450 from Claxton to Valdosta 150 from Millen to Savannah 300 from Millen to Perry Total Profit = 12,775

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Ragsdale 5-16 by Dave Neal A company has 3 warehouses that supply 4 stores with a given product. Each warehouse has 30 units of the product (Total Supply = 90 units). Stores 1,2,3,4 require 20,25,30,35 units respectively (Total Demand = 110 units). PROBLEM: Determine least expensive shipping plan to fill store demand.

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Ragsdale 5-16 by Dave Neal

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Ragsdale 5-16 by Dave Neal Initial Problem Set-Up Type of Problem: Transportation Objective Function: Minimize Shipping Cost MIN: 5 X11 + 4 X12 + 6 X13 + 5 X14 + 3 X21 + 6 X22 + 4 X23 + 4 X24 + 4 X31 + 3 X32 + 3 X33 + 2 X34 Constraints: -X11 - X12 - X13 - X14 = -30 -X21 - X22 - X23 - X24 = -30 -X31 - X32 - X33 - X34 = -30 +X11 + X21 + X31 <= +20 +X12 + X22 + X32 <= +25 +X13 + X23+ X33 <= +30 +X14 + X24+ X34 <= +35 Xij >= 0 NOTE: SUPPLY < DEMAND: 90 < 110 For minimum cost network flow problems where total supply

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Ragsdale 5-16 by Dave Neal Initial Excel Settings

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Ragsdale 5-16 by Dave Neal Results

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Ragsdale 5-16 by Dave Neal Revised Excel Settings No shipments between warehouse 1, store 2 and warehouse 2, store 3. Added 2 constraints to solve the modified problem. X12, X23 = 0

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Ragsdale 5-16 by Dave Neal Revised Results

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Ragsdale Chap 5-19 by Deyanira Pena Net flow Toulon 3 Doha 1 Port 6 Suez 5 Damietta 7 Rotterdam 2 Palermo 4 +15 +6 +9 - -30 $.16 $.35 $.20 $.15 $1.35 $.19 $.25 $.23 $.20$1.40 $1.20 $.27

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Ragsdale Chap 5-19 by Deyanira Pena Lp Model Xij = the number of barrels shipped from node i to node j X12 = the number of barrels shipped from node 1(Doha) to node 2 (Rotterdam) X13 the number of barrels shipped from node 1(Doha) to node 3 (Toulon) X14 the number of barrels shipped from node 1(Doha) to node 4 (Palermo) X15 the number of barrels shipped from node 1(Doha) to node 5(Suez) X56 the number of barrels shipped from node 5(Suez) to node 6 (Port) X57 the number of barrels shipped from node 5(Suez) to node 7(Damietta) X62 the number of barrels shipped from node 6(Port) to node 2 (Rotterdam)

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Ragsdale Chap 5-19 by Deyanira Pena Lp Model Cont. X63 the number of barrels shipped from node 6 (Port) to node 3 (Toulon) X64 the number of barrels shipped from node 6(Port) to node 4 (Palermo) X72 the number of barrels shipped from node 7(Damietta) to node 2 (Rotterdam) X73 the number of barrels shipped from node 7(Damietta) to node 3 (Toulon) X74 the number of barrels shipped from node 7(Damietta) to node 4 (Palermo) Min: + 1.20X12 + 1.40X13 + 1.35X14 +.20X56 +.27X62 +.23X63 +.19X64 +.35X15 +.16X57 +.25X72+.20X73 +.15X74

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Ragsdale Chap 5-19 by Deyanira Pena Lp Model Cont. Constraints Subject To - X12 - X13 -X14 >= -3000000 + X12 + X13 + X14 >= 2500000 - X15 + X56 + X62 >= 6000000 - X15 + X56 + X63 >= 1500000 - X15 + X56 + X62 >= 9000000 - X15 – X57 + X72 + X73 + X74 >= 15000000

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Ragsdale Chap 5-19 by Deyanira Pena Spreadsheet

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Ragsdale Chap 5-19 by Deyanira Pena Optimal Solution

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Ragsdale 6-9 by Dave Neal Health Care Systems of Florida planning to build emergency-care clinics. Management divided area into 7 regions. All 7 regions must be served by at least 1 of the 5 possible facility sites. PROBLEM: Determine which sites to select that will result in the least cost while providing convenient service to all locations.

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Ragsdale 6-9 by Dave Neal Initial Problem Set-Up Type of Problem: Integer Linear Programming Model / Capital Budgeting Problem Objective Function: Minimize cost while providing convenient service to all locations. MIN: 450X1 + 650X2 + 550X3 + 500X4 + 525X5 Constraints: X1 + X3 >= 1 X1 + X2 + X4 + X5 >= 1 X2 +X4 > = 1 X3 +X5 > = 1 X1 +X2 > = 1 X3 +X5 > = 1 X4 +X5 > = 1 Xi must be BINARY, i = 1,2,3,4,5 Xi = 1, if building site i is selected Xi = 0, otherwise

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Ragsdale 6-9 by Dave Neal Initial Excel Settings

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Ragsdale 6-9 by Dave Neal

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Ragsdale Chap 6-12 by Deyanira Pena Xi= { 1,if investment I is selected i= 1,2,…,5 0,otherwise Max : 30X1 + 30X2 +30X3+ 30X4+ 30X5 Subject to: 35X1 + 16X2 +125X3+ 25X4+40X5 + 5X6 30 } year 1 investment value 37X1 + 17X2 +130X3+ 27X4+43X5 + 7X6 30 } year 2 investment value 39X1 + 18X2 +136X3+ 29X4+46X5 + 8X6 30 } year 3 investment value 42X1+19X2 +139X3+ 30X4+50X5 +10X6 30 } year 4 investment value 45X1 +20X2 +144X3+ 33X4+52X5 + 11X6 30 } year 5 investment value

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Ragsdale Chap 6-12 by Deyanira Pena Spreadsheet Model

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Ragsdale Chap 6-12 by Deyanira Pena optimal solution

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Ragsdale 6-15 by Rosele Lynn Problem: Where should a manufacturer build its new plants if it wants to be closer to its main supply customers X,Y,Z? Decision variables: Of the 5 alternative plants, which plants should the manufacturer build? P1= plant 1, 1 is selected, 0 if it is not selected P2=plant 2, 1 is selected, 0 if it is not selected P3=plant 3, 1 is selected, 0 if it is not selected P4=plant 4, 1 is selected, 0 if it is not selected P5=plant 5, 1 is selected, 0 if it is not selected

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Ragsdale 6-15 by Rosele Lynn Objective Function: Which plant should be built in order to satisfy customer demand at a minimum cost? MIN: 35 P1X + 30 P1Y + 45 P1Z + 45 P2X + 40 P2Y + 50 P2Z + 70 P3X + 65 P3Y + 50 P3Z + 20 P4X + 45 P4Y + 25 P4Z + 65 P5X + 45 P5Y + 45 P5Z + 1000*(1,325 Y1 + 1,100 Y2 + 1,500 Y3 + 1,200 Y4 + 1,400 Y5) Constraints: Decision to build is Binary, Yi = binary Production Capacity for Plants 1,2,3,4,5 are as follows: P1X + P1Y + P1Z < 40,000 Y1 P2X + P2Y + P2Z <30,000 Y2 P3X + P3Y + P3Z < 50,000 Y3 P4X + P4Y + P4Z <20,000 Y4 P5X + P5Y + P5Z <40,000 Y5 Expected Demand: 40,000 from Customer X, 25,000 from Customer Y, 35,000 from Customer Z P1X + P2X + P3X + P4X + P5X > 40,000 P1Y + P2Y + P3Y + P4Y + P5Y >25,000 P1Z + P2Z + P3Z + P4Z + P5Z >35,000

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Ragsdale 6-15 by Rosele Lynn

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Ragsdale 6-15 by Rosele Lynn Solver Parameters

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The optimal solution is to build plant 1, 4, and 5. Here all constraints are satisfied and the binary is 1 (yes, to build). Ragsdale 6-15 by Rosele Lynn Plant Location Problem

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Ragsdale 6-18 by Reynald Decision Variables X1 = the barrels to buy from TX X2 = the barrels to buy from OK X3 = the barrels to buy from PA X4 = the barrels to buy from AL Y1,Y2,Y3,Y4 = 1 if X >0 and 0 otherwise Objective Function 22X 1 + 21X 2 + 22X 3 + 24X 4 + 1500Y 1 + 1700Y 2 + 1500Y 3 + 1400Y 4 Constraints Numbers to be produced 2X 11 + 1.8X 21 + 2.3X 31 + 2.1X 41 >= 750 2.8X 12 + 2.3X 22 + 2.2X 32 + 2.6X 42 >= 800 1.70X 13 + 1.75X 23 + 1.6X 33 + 1.9X 43 >= 1000 2.4X 14 + 1.90X 24 + 2.6X 34 + 2.4X 44 >= 300

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Ragsdale 6-18 by Reynald Constraints (cont’) Minimum required X 1 – 500Y 1 >= 0 X 2 – 500Y 2 >= 0 X 3 – 500Y 3 >= 0 X 4 – 500Y 4 >= 0 Maximum X 1 – 1500Y 1 <= 0 X 2 – 2000Y 2 <=0 X 3 – 1500Y 3 <= 0 X 4 – 1800Y 4 <=0

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Ragsdale 6-18 by Reynald Excel

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Ragsdale 6-18 by Reynald Solver

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Ragsdale 6-18 by Reynald Conclusion The company should purchase 1316 barrels from Alabama. The total cost will be $31,671

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Dielman 4-1 by Reynald VarCoeStd DevT statP val Inter51.7221.702.380.026 Paper0.950.127.900.000 Mach2.470.475.310.000 Over0.050.530.090.92 Labor-0.050.04-1.260.223 Standard Error = 11.0756; R-Sq = 99.9%; R-Sq(avg) = 99.9% sourceDFSum SqMean SQF statP val Regression422714235678564629.170.000 Error222699123 Total262274122 Analysis of Variance

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Dielman 4-1 by Reynald Cont. A) What is the equation? COST = 51.72 + 0.95Paper + 2.47Machine +0.05Overhead – 0.05Labor B) Conduct an F test. Decision Rule: Reject Ho if F> F(0.05; 4, 22) = 2.82 Test Stat: F = 4629.17 Reject Ho. At least one of the coefficients is not equal to zero C) Find 95% confidence interval estimate 2.47, 2.47 +- ( 2.074 )( 0.47 ) D) Conduct two-tailed test procedure Critical Value: t(0.025, 22) = 2.074 Test Statistic: t = -0.42 Decision: Do not reject Ho. The true marginal cost is 1. E) What percentage have been explained? 99.9% F) What is the adjusted R squared? 99.9% G) What action might be taken. This information can be use to reduce cost. It shows the influence of one variable

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The estimated Regression equation is: FUELCON = 916 - 218 DRIVERS - 0.00078 HWYMILES - 3.69 GASTAX - 0.00549 INCOME. Using a 5% level of significance. Gas tax and drivers are the most significant factors in this regression model. From the regression: S = 56.2806 R-Sq = 44.4% R-Sq(adj) = 39.6% PRESS = 190252 R-Sq(pred) = 27.40% Income and Highway Miles appear to be unnecessary in the regression. Their factor is very insignificant in the equation and the model. No Variables where omitted from the regression. Dielman 4.11 by Thamer

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Cont.Dielman 4.11 by Thamer Regression

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